NE STUD 180 Midterm: ne180-fa2010-mt2-Morse-soln

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8 Jan 2019
School
Professor
NE 180
Midterm II
Fall Semester 2009
Solutions
November 3, 2010
1. a.
Ti(0.9a) = 15 ×1031(0.9)2)0.3= 9114.17eV
ni(0.9a) = 1.0×1020 m3= 1.0×1014 cm3
i=2.09 ·107(mi/mp)1/2T3/2
i
ln Λ = 1.64 ×1012 cm3s
τi= 0.016 s
ωci =qB
mi
=1.53 ×108s1
ωciτi= 2.5187 ×106
κi
= 2.0nTiτi
mi(ωciτi)2= 1.8093 ×1017m1s1
1. b.
dT
dr = (0.3)2r
a2 1r
a2!0.7
=12951 eV m1
1
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Document Summary

Ti(0. 9 a) = 15 103(cid:16)1 (0. 9)2)(cid:17)0. 3. = 9114. 17ev ni(0. 9 a) = 1. 0 1020 m 3 = 1. 0 1014 cm 3 n i = = 2. 0 nti i mi( ci i)2 = 1. 8093 1017m 1 s 1: b. dt dr. Total ion heat flux = area q = 159. 86 kw: c. q = 1/(1 (2/3)(r/a)2) = 2. 17391. Neoclassical factor = qneo = 2q2 3/2 = 57. 5218. Since q = qneo dt (r) dr. At center, ti te = 15 10 = 5. 0 kev. E = 2. 0 1010t 3/2 e /ne with te in ev and mks density. = 2. 8 10 8zt 3/2 e(kev ) Then pe = 2 c 1 ne = 2. 59959 1031m 3 crit = 2. 87 1017 s 1 and, since 2 pe = nee2/(me 0), we have: b. = nmi = 108. 316 g cm 3 , / liq = 433. 264: c. Then fb = r/(6 + r) = 0. 142857 and y = 3. 39 1011j g 1.

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